A 5 digit number is formed by using the digits 0,1,2,3,4 and 5 without repetition. The probability that the number is divisible by 6 is:
A 5 digit number is formed by using the digits 0,1,2,3,4 and 5 without repetition. The probability that the number is divisible by 6 is:
The correct option is C 18%
Total 55 digit numbers formed by the given digits=5∗5!=6005∗5!=600
For divisibility by 6, the numbers must be divisible by both 2 and 3.
Only numbers formed by digits (i) (0,1,2,4,5) and (ii) (1,2,3,4,5) are divisible by 3.
(i) CASE 1: ( 0,1,2,4,5) For divisibility by 2, _ _ _ _ _
(a) if 0 is placed at the end, then number of ways=4!=24 (b) if 0 is not placed in the end, then there are two ways to select the last digit(2,4). Number of ways=2*3*3!=362C1∗3C1∗3!=36 For this case, total number of ways=24+36=60
(ii) CASE 2: ( 1,2,3,4,5) Again, the last number can be chosen in 22C1 ways and the remaining 4 numbers can permute in 4!4! ways. The required number of ways for this case= 2∗4!=482∗4!=48
Adding the number of ways in the two cases, 5 digit numbers that are divisible by 6= 108
Required Probability = 108600=108600=0.18==18%
Total 55 digit numbers formed by the given digits=5∗5!=6005∗5!=600
For divisibility by 6, the numbers must be divisible by both 2 and 3.
Only numbers formed by digits (i) (0,1,2,4,5) and (ii) (1,2,3,4,5) are divisible by 3.
(i) CASE 1: ( 0,1,2,4,5) For divisibility by 2, _ _ _ _ _
(a) if 0 is placed at the end, then number of ways=4!=24 (b) if 0 is not placed in the end, then there are two ways to select the last digit(2,4). Number of ways=2*3*3!=362C1∗3C1∗3!=36 For this case, total number of ways=24+36=60
(ii) CASE 2: ( 1,2,3,4,5) Again, the last number can be chosen in 22C1 ways and the remaining 4 numbers can permute in 4!4! ways. The required number of ways for this case= 2∗4!=482∗4!=48
Adding the number of ways in the two cases, 5 digit numbers that are divisible by 6= 108
Required Probability = 108600=108600=0.18==18%
